What is the derivative of (2t)/(t^2+4) ?

The derivative of (2t)/(t^2+4) is (2(-t^2+4))/((t^2+4)^2)

What is the first derivative of (2t)/(t^2+4) ?

The first derivative of (2t)/(t^2+4) is (2(-t^2+4))/((t^2+4)^2)

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derivative of

\frac{2 t}{t ^{2} + 4}

\frac{2 ( - t ^{2} + 4 )}{( t ^{2} + 4 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{2 t}{t ^{2} + 4})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d t} (\frac{t}{t ^{2} + 4})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 2 \cdot \frac{\frac{d t}{d t} ( t ^{2} + 4 ) - \frac{d}{d t} ( t ^{2} + 4 ) t}{( t ^{2} + 4 ) ^{2}}

\frac{d t}{d t} = 1

\frac{d}{d t} (t^{2} + 4) = 2 t

= 2 \cdot \frac{1 \cdot ( t ^{2} + 4 ) - 2 tt}{( t ^{2} + 4 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( t ^{2} + 4 ) - 2 tt}{( t ^{2} + 4 ) ^{2}} : \frac{2 ( - t ^{2} + 4 )}{( t ^{2} + 4 ) ^{2}}

= \frac{2 ( - t ^{2} + 4 )}{( t ^{2} + 4 ) ^{2}}

(5, - 3), (0, - 9)

\frac{\partial}{\partial y} (e^{3 x} \cdot tan (y))

\frac{d}{d x} ((x^{2} + 4)^{2} (2 x^{3} - 1)^{3})

f (x) = \frac{2}{1 + e ^{- x}}

\int 7 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

What is the derivative of (2t)/(t^2+4) ?
The derivative of (2t)/(t^2+4) is (2(-t^2+4))/((t^2+4)^2)
What is the first derivative of (2t)/(t^2+4) ?
The first derivative of (2t)/(t^2+4) is (2(-t^2+4))/((t^2+4)^2)